$\newcommand{\R}{\mathbb R}$Consider the following construction. For real $u$, let
\begin{equation}
f(u):=\frac{2u^2}{1+u^4},
\end{equation}
so that the function $f\colon\R\to\R$ is continuous, $0\le f\le1=f(\pm1)$, and $f(u)\to0=f(0)$ as $|u|\to\infty$.

Let $Q$ be any countable dense subset of $\R$. Let $(w_q)_{q\in Q}$ be any family of (strictly) positive real numbers such that $\sum_{q\in Q}w_q<\infty$.
Finally, for each natural $n$ and all real $x$, let
\begin{equation}
g_n(x):=\sum_{q\in Q}w_q\, f(n(x-q)).
\end{equation}

The latter series converges uniformly in $x$, and hence the function $g_n$ is continuous. Moreover, by dominated convergence, $g_n\to0$ pointwise (as $n\to\infty$). Take now any nonempty open interval $I\subset\R$. Then $r\in I$ for some $r\in Q$. Moreover, $\{r+1/n,r-1/n\}\cap I\ne\emptyset$ eventually -- that is, for all large enough $n$ (depending on $I$). So, \begin{equation} \liminf_n\,\sup_I g_n\ge w_r \liminf_n f(n(r\pm1/n-r))=w_r>0. \end{equation}

Thus, we have a sequence of continuous functions $g_n$ converging to $0$ pointwise on $\R$ while for any nonempty open interval $I\subset\R$ we have $\liminf\limits_n\,\sup\limits_I g_n>0$.

This brings us to

Question:Does there exist a sequence of continuous functions $g_n\colon\R\to\R$ converging to $0$ pointwise on $\R$ such that for any nonempty open interval $I\subset\R$ we have $\liminf\limits_n\,\sup\limits_I g_n\ge1$?

(Of course, here we can replace $\ge1$ by $=1$.)